Hi,

I'm trying to get a circuit to work which gets activated by the output of a device which sources 5V @1mA, to drive a relay. Once the 5V is removed, I would like to hold the relay activated for a short period of time, 2-3 seconds. I have tried the circuit below, but when the 5V is removed, the relay never de-activates. Any help appreciated.

 

The input changing as the b) fig of below.

The original function is set the delay time when switch on and off.
Now you just changing the delay time when the signal is off, you could refer to the circuit as below.

You can try the values and combine them together by yourself.

 

Hi Scott, Thank you for the reply. Not sure I understand though. If I use your delay design, can I still use my 12V relay? I would need to switch out the +5 to +12 on the relay, but use a voltage divider for the collector of Q1?

 

You could also use an edge triggered 555 etc to get the delay,and why use 2 transistors to drive your relay when one would do?

 

Hi Scott, Thank you for the reply. Not sure I understand though. If I use your delay design, can I still use my 12V relay? I would need to switch out the +5 to +12 on the relay, but use a voltage divider for the collector of Q1?

The b) of the first circuit diagram:
The input signal is came from your 5V, and the V1 is +12V, RL is no need, Vo is connecting to R1(220K) of the second circuit, so the function of Q2 as the sw1 of the second circuit.

The second circuit diagram:
Changing 5V to 12V
Because you just need the delay time when 5V signal is removed, so R1(220K) is no need a high value, it can be 1K or as your R2(560Ω).

Reduce the value of R2(680K) will reduce the delay time, you can try a Appropriate value for your need.